Given the head of a linked list, remove the nth node from the end of the list and return its head.
Example 1:

Input: head = [1,2,3,4,5], n = 2 Output: [1,2,3,5]
Example 2:
Input: head = [1], n = 1 Output: []
Example 3:
Input: head = [1,2], n = 1 Output: [1]
Constraints:
- The number of nodes in the list is
sz. 1 <= sz <= 300 <= Node.val <= 1001 <= n <= sz
Follow up: Could you do this in one pass?
Approach 01: Two Pointers (One Pass)
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* dummy = new ListNode(0, head);
ListNode* first = dummy;
ListNode* second = dummy;
for (int i = 0; i <= n; i++) {
first = first->next;
}
while (first != nullptr) {
first = first->next;
second = second->next;
}
second->next = second->next->next;
return dummy->next;
}
};
Time Complexity:
- O(L): where L is the number of nodes in the list. One pass.
Space Complexity:
- O(1): Constant space used.
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
dummy = ListNode(0, head)
first = dummy
second = dummy
for i in range(n + 1):
first = first.next
while first:
first = first.next
second = second.next
second.next = second.next.next
return dummy.next
Time Complexity:
- O(L): where L is the number of nodes in the list. One pass.
Space Complexity:
- O(1): Constant space used.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode first = dummy;
ListNode second = dummy;
for (int i = 1; i <= n + 1; i++) {
first = first.next;
}
while (first != null) {
first = first.next;
second = second.next;
}
second.next = second.next.next;
return dummy.next;
}
}
Time Complexity:
- O(L): where L is the number of nodes in the list. One pass.
Space Complexity:
- O(1): Constant space used.